I started preparing for an interview and came across this problem:
An array of integers is given
Now calculate the sum of Hamming distances of all pairs of integers in the array in their binary representation.
Example:
given {1,2,3} or {001,010,011} (used 3 bits just to simplify)
result= HD(001,010)+HD(001,011)+HD(010,011)= 2+1+1=4;
The only optimization, from a purely brute force solution, I know I can use here, is in the individual calculation of Hamming Distance as seen here:
int hamming_distance(unsigned x, unsigned y)
{
int dist;
unsigned val;
dist = 0;
val = x ^ y; // XOR
// Count the number of bits set
while (val != 0)
{
// A bit is set, so increment the count and clear the bit
dist++;
val &= val - 1;
}
// Return the number of differing bits
return dist;
}
What's the best way to go about solving this problem?
Here is my C++ implementation, with O(n) complexity and O(1) space.
int sumOfHammingDistance(vector<unsigned>& nums) {
int n = sizeof(unsigned) * 8;
int len = nums.size();
vector<int> countOfOnes(n, 0);
for (int i = 0; i < len; i++) {
for (int j = 0; j < n; j++) {
countOfOnes[j] += (nums[i] >> j) & 1;
}
}
int sum = 0;
for (int count: countOfOnes) {
sum += count * (len - count);
}
return sum;
}
You can consider the bit-positions separately. That gives you 32 (or some other number) of easier problems, where you still have to calculate the sum of all pairs of hamming distances, except now it's over 1-bit numbers.
The hamming distance between two 1-bit numbers is their XOR.
And now it has become the easiest case of this problem - it's already split per bit.
So to reiterate the answer to that question, you take a bit position, count the number of 0's and the number of 1's, multiply those to get the contribution of this bit position. Sum those for all bit positions. It's even simpler than the linked problem, because the weight of the contribution of every bit is 1 in this problem.
This might be a bit cryptic title but I have a very specific problem. First my current setup
Namely in my card simulator I deal 32 cards to 4 players in sets of 8. So 8 cards per player.
With the 4 standard suits (spades, harts , etc)
My current implementation cycles threw all combinations of 8 out of 32
witch gives me a large number of possibilities.
Namely the first player can have 10518300 different hands be dealt.
The second can then be dealt 735471 different hands.
The third player then 12870 different hands.
and finally the fourth can have only 1
giving me a grand total of 9.9561092e+16 different unique ways to deal a deck of 32 cards to 4 players. if the order of cards doesn’t matter.
On a 4 Ghz processor even with 1 tick per possibility it would take me half a year.
However I would like to simplify this dealing of cards by making the exchange of diamonds, harts and spades. Meaning that dealing of 8 harts to player 1 is equivalent to dealing 8 spades. (note that this doesn’t apply to clubs)
I am looking for a way to generate this. Because this will cut down the possibilities of the first hand by at least a factor of 6. My current implementation is in c++.
But feel free to answer in a different Languages
/** http://stackoverflow.com/a/9331125 */
unsigned cjasMain::nChoosek( unsigned n, unsigned k )
{
//assert(k < n);
if (k > n) return 0;
if (k * 2 > n) k = n-k;
if (k == 0) return 1;
int result = n;
for( int i = 2; i <= k; ++i ) {
result *= (n-i+1);
result /= i;
}
return result;
}
/** [combination c n p x]
* get the [x]th lexicographically ordered set of [r] elements in [n]
* output is in [c], and should be sizeof(int)*[r]
* http://stackoverflow.com/a/794 */
void cjasMain::Combination(int8_t* c,unsigned n,unsigned r, unsigned x){
++x;
assert(x>0);
int i,p,k = 0;
for(i=0;i<r-1;i++){
c[i] = (i != 0) ? c[i-1] : 0;
do {
c[i]++;
p = nChoosek(n-c[i],r-(i+1));
k = k + p;
} while(k < x);
k = k - p;
}
c[r-1] = c[r-2] + x - k;
}
/**http://stackoverflow.com/a/9430993 */
template <unsigned n,std::size_t r>
void cjasMain::Combinations()
{
static_assert(n>=r,"error n needs to be larger then r");
std::vector<bool> v(n);
std::fill(v.begin() + r, v.end(), true);
do
{
for (int i = 0; i < n; ++i)
{
if (!v[i])
{
COUT << (i+1) << " ";
}
}
static int j=0;
COUT <<'\t'<< j++<< "\n";
}
while (std::next_permutation(v.begin(), v.end()));
return;
}
A requirement is that from lexicographical number I can get back the original array.
Even the slightest optimization can help my monto carol simulation I hope.
i have created the recursive call tree by applying brute force technique but when i give this algorithm 100 values it takes trillion of years to compute..
what you guys suggest me to do that it runs fast by giving 100 values
here is what i have done so far
function fib(n) {
if (n =< 1) {
return n;
} else {
return fib(n - 1) + fib(n - 2);
}
}
You can do it also with a loop:
int a = 1;
int b = 1;
for(int i = 2; i < 100; i++){
int temp = a + b;
a = b;
b = temp;
}
System.out.println("Fib 100 is: "+b);
The runtime is linear and avoids the overhead caused by the recursive calls.
EDIT: Please note that the result is wrong. Since Fib(100) is bigger than Integer.MAX_VALUE you have to use BigInteger or similar to get the correct output but the "logic" will stay the same.
You could have a "cache", where you save already computed Fibonacci numbers. Every time you try to compute
fib(n-1) /* or */ fib(n-2) ;
You would first look into your array of already computed numbers. If it's there, you save a whole lot of time.
So every time you do compute a fibonacci number, save it into your array or list, at the corresponding index.
function fib(n)
{
if (n =< 1)
{
return n;
}
if(fiboList[n] != defaultValue)
{
return fiboList[n];
}
else
{
int fibo = fib(n-1) + fib(n-2);
fiboList[n] = fibo;
return fibo;
}
}
You can also do it by dynamic programming:
def fibo(n):
dp = [0,1] + ([0]*n)
def dpfib(n):
return dp[n-1] + dp[n-2]
for i in range(2,n+2):
dp[i] = dpfib(i)
return dp[n]
I did make algorithm for creating Bézier Curve with Objective-C and Cocos2D. Here is my code
-(int)factorial:(int)x{
int sum=1;
int i;
if(x == 0){
return 1;
}else{
for(i=1;i<x;i++){
sum = sum*i;
}
return sum;
}
}
-(int)binomialCoefficient:(int)n:(int)i{
int sum;
//NSLog([NSString stringWithFormat:#"fac n-i=%f\n", fach] );
sum = [self factorial:n]/([self factorial:i]*[self factorial:(n-i)]);
return sum;
}
-(float)convertT:(int)t{
return t*(0.001);
}
-(float)power:(float)a:(int)b{
int i;
float hasil=1;
for(i=0;i<b;i++){
hasil = hasil*a;
}
return hasil;
}
-(float)bernstein:(float)t:(int)n:(int)i{
float sum = 0;
sum = [self binomialCoefficient:n:i]*[self power:t :i]*[self power:(1-t) :(n-i)];
//NSLog([NSString stringWithFormat:#"yeah"]);
return sum;
}
and for implementation you just put an array of x and y and access it. For example to draw a single dot in control curve I did it like this
float myPx = px[i];
float myPy = py[i];
posx = posx+([self bernstein:theT :banyak-1 :i]*myPx);
posy = posy+([self bernstein:theT :banyak-1 :i]*myPy);
Yes, this code doesn't give the perfect nice line, but I try to draw it dot by dot.
It works well, but the problem arise when I try to use 3 dots. The middle dot for curving the lines didn't behave like what I expected. For example if I put 3 dots in these coordinates:
a(100,200)
b(250,250)
c(500,200)
It didn't curving up but curving down. If I want to put it straight I have to put it all the way higher.
Am I do it wrong in syntax or data types? Or is it just my algorithm?
Thanks in advance
Best Regards
(sorry for my bad english)
The factorial loop should be
for ( i = 1 ; i <= x ; i++ )
instead of
for ( i = 1 ; i < x ; i++ )
How many possible combinations of the variables a,b,c,d,e are possible if I know that:
a+b+c+d+e = 500
and that they are all integers and >= 0, so I know they are finite.
#Torlack, #Jason Cohen: Recursion is a bad idea here, because there are "overlapping subproblems." I.e., If you choose a as 1 and b as 2, then you have 3 variables left that should add up to 497; you arrive at the same subproblem by choosing a as 2 and b as 1. (The number of such coincidences explodes as the numbers grow.)
The traditional way to attack such a problem is dynamic programming: build a table bottom-up of the solutions to the sub-problems (starting with "how many combinations of 1 variable add up to 0?") then building up through iteration (the solution to "how many combinations of n variables add up to k?" is the sum of the solutions to "how many combinations of n-1 variables add up to j?" with 0 <= j <= k).
public static long getCombos( int n, int sum ) {
// tab[i][j] is how many combinations of (i+1) vars add up to j
long[][] tab = new long[n][sum+1];
// # of combos of 1 var for any sum is 1
for( int j=0; j < tab[0].length; ++j ) {
tab[0][j] = 1;
}
for( int i=1; i < tab.length; ++i ) {
for( int j=0; j < tab[i].length; ++j ) {
// # combos of (i+1) vars adding up to j is the sum of the #
// of combos of i vars adding up to k, for all 0 <= k <= j
// (choosing i vars forces the choice of the (i+1)st).
tab[i][j] = 0;
for( int k=0; k <= j; ++k ) {
tab[i][j] += tab[i-1][k];
}
}
}
return tab[n-1][sum];
}
$ time java Combos
2656615626
real 0m0.151s
user 0m0.120s
sys 0m0.012s
The answer to your question is 2656615626.
Here's the code that generates the answer:
public static long getNumCombinations( int summands, int sum )
{
if ( summands <= 1 )
return 1;
long combos = 0;
for ( int a = 0 ; a <= sum ; a++ )
combos += getNumCombinations( summands-1, sum-a );
return combos;
}
In your case, summands is 5 and sum is 500.
Note that this code is slow. If you need speed, cache the results from summand,sum pairs.
I'm assuming you want numbers >=0. If you want >0, replace the loop initialization with a = 1 and the loop condition with a < sum. I'm also assuming you want permutations (e.g. 1+2+3+4+5 plus 2+1+3+4+5 etc). You could change the for-loop if you wanted a >= b >= c >= d >= e.
I solved this problem for my dad a couple months ago...extend for your use. These tend to be one time problems so I didn't go for the most reusable...
a+b+c+d = sum
i = number of combinations
for (a=0;a<=sum;a++)
{
for (b = 0; b <= (sum - a); b++)
{
for (c = 0; c <= (sum - a - b); c++)
{
//d = sum - a - b - c;
i++
}
}
}
This would actually be a good question to ask on an interview as it is simple enough that you could write up on a white board, but complex enough that it might trip someone up if they don't think carefully enough about it. Also, you can also for two different answers which cause the implementation to be quite different.
Order Matters
If the order matters then any solution needs to allow for zero to appear for any of the variables; thus, the most straight forward solution would be as follows:
public class Combos {
public static void main() {
long counter = 0;
for (int a = 0; a <= 500; a++) {
for (int b = 0; b <= (500 - a); b++) {
for (int c = 0; c <= (500 - a - b); c++) {
for (int d = 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2656615626.
Order Does Not Matter
If the order does not matter then the solution is not that much harder as you just need to make sure that zero isn't possible unless sum has already been found.
public class Combos {
public static void main() {
long counter = 0;
for (int a = 1; a <= 500; a++) {
for (int b = (a != 500) ? 1 : 0; b <= (500 - a); b++) {
for (int c = (a + b != 500) ? 1 : 0; c <= (500 - a - b); c++) {
for (int d = (a + b + c != 500) ? 1 : 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2573155876.
One way of looking at the problem is as follows:
First, a can be any value from 0 to 500. Then if follows that b+c+d+e = 500-a. This reduces the problem by one variable. Recurse until done.
For example, if a is 500, then b+c+d+e=0 which means that for the case of a = 500, there is only one combination of values for b,c,d and e.
If a is 300, then b+c+d+e=200, which is in fact the same problem as the original problem, just reduced by one variable.
Note: As Chris points out, this is a horrible way of actually trying to solve the problem.
link text
If they are a real numbers then infinite ... otherwise it is a bit trickier.
(OK, for any computer representation of a real number there would be a finite count ... but it would be big!)
It has general formulae, if
a + b + c + d = N
Then number of non-negative integral solution will be C(N + number_of_variable - 1, N)
#Chris Conway answer is correct. I have tested with a simple code that is suitable for smaller sums.
long counter = 0;
int sum=25;
for (int a = 0; a <= sum; a++) {
for (int b = 0; b <= sum ; b++) {
for (int c = 0; c <= sum; c++) {
for (int d = 0; d <= sum; d++) {
for (int e = 0; e <= sum; e++) {
if ((a+b+c+d+e)==sum) counter=counter+1L;
}
}
}
}
}
System.out.println("counter e "+counter);
The answer in math is 504!/(500! * 4!).
Formally, for x1+x2+...xk=n, the number of combination of nonnegative number x1,...xk is the binomial coefficient: (k-1)-combination out of a set containing (n+k-1) elements.
The intuition is to choose (k-1) points from (n+k-1) points and use the number of points between two chosen points to represent a number in x1,..xk.
Sorry about the poor math edition for my fist time answering Stack Overflow.
Just a test for code block
Just a test for code block
Just a test for code block
Including negatives? Infinite.
Including only positives? In this case they wouldn't be called "integers", but "naturals", instead. In this case... I can't really solve this, I wish I could, but my math is too rusty. There is probably some crazy integral way to solve this. I can give some pointers for the math skilled around.
being x the end result,
the range of a would be from 0 to x,
the range of b would be from 0 to (x - a),
the range of c would be from 0 to (x - a - b),
and so forth until the e.
The answer is the sum of all those possibilities.
I am trying to find some more direct formula on Google, but I am really low on my Google-Fu today...